# Question Video: Finding the First Derivative of a Composition of Radical and Trigonometric Functions Using the Quotient Rule Mathematics • Higher Education

Determine the derivative of the function π (π‘) = β((βsin π‘ + 7)/(βcos π‘ + 7)).

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### Video Transcript

Determine the derivative of the function π  of π‘ equals the square root of negative sin π‘ plus seven over negative cos π‘ plus seven.

Now, we can see straightaway in this question that we have a quotient. So, weβre going to need to apply the quotient rule at some point. But are we going to need to do anything else? Well, we donβt just have this quotient. We have the square root of this quotient, which means we have a composite function. And so, weβre also going to need to apply the chain rule. Weβll begin by allowing π’ to be that quotient underneath the square root. π’ is equal to negative sin π‘ plus seven over negative cos π‘ plus seven. Then, π  becomes a function of π’. Itβs equal to the square root of π’, which we can express using index notation as π’ to the power of one-half.

The chain rule, using the letters π , π‘, and π’ as we have in this question, tells us that the derivative of π  with the respect to π‘ is equal to dπ  by dπ’ multiplied by dπ’ by dπ‘. Applying the power rule, we see that dπ  by dπ’ is equal to one-half π’ to the power of negative a half. But in order to find dπ’ by dπ‘, weβre going to need to apply the quotient rule.

Weβll define π to be the function in the numerator, thatβs negative sin π‘ plus seven, and π to be the function in the denominator, negative cos π‘ plus seven. In order to find the derivatives of these two functions, we need to recall how we differentiate sin and cos. There is a helpful little cycle that we can remember. The derivative of sin π‘ is cos π‘. The derivative of cos π‘ is negative sin π‘. The derivative of negative sin π‘ is negative cos π‘. And the derivative of negative cos π‘ is sin π‘. And then we go around the cycle again.

Remembering that the derivative of a constant is just zero, we have that π prime is equal to negative cos π‘ and π prime is equal to sin π‘. Now, we can substitute into the quotient rule to find dπ’ by dπ‘. Itβs π, thatβs negative cos π‘ plus seven, multiplied by π prime, negative cos π‘, minus π, thatβs negative sin π‘ plus seven, multiplied by π prime, thatβs sin π‘, all over π squared. Now, we need to do some simplification. So, weβll expand the parentheses in the numerator, which gives cos squared π‘ minus seven cos π‘ plus sin squared π‘ minus seven sin π‘ over negative cos π‘ plus seven squared.

We can recall, at this point, one of our trigonometric identities, cos squared π‘ plus sin squared π‘ is equal to one. So, this simplifies to one minus seven cos π‘ minus seven sin π‘ over negative cos π‘ plus seven squared. Now that we found both dπ’ by dπ‘ and dπ  by dπ’, we can substitute into the chain rule. We have then that dπ  by dπ‘ is equal to dπ  by dπ’, thatβs a half π’ to the power of negative a half, multiplied by dπ’ by dπ‘, thatβs one minus seven cos π‘ minus seven sin π‘ over negative cos π‘ plus seven squared.

Remember, though, that dπ  by dπ‘ must be in terms of π‘ only. So, we need to reverse our substitution. We have negative sin π‘ plus seven over negative cos π‘ plus seven to the power of negative a half multiplied by one minus seven cos π‘ minus seven sin π‘ over two multiplied by negative cos π‘ plus seven squared. That power of negative a half means a reciprocal, so we can deal with that by inverting the fraction. And the first part becomes negative cos π‘ plus seven over negative sin π‘ plus seven to the power of positive one-half.

We can then simplify the powers. We have negative cos π‘ plus seven to the power of a half in the numerator and then negative cos π‘ plus seven to the power of two in the denominator. Which will lead to a power of negative three over two overall. Thatβs a power of three over two in the denominator. This leads us to one minus seven cos π‘ minus seven sin π‘ in the numerator. And in the denominator two times the square root of negative sin π‘ plus seven. Thatβs negative sin π‘ plus seven to the power of a half multiplied by negative cos π‘ plus seven to the power of three over two.

In this question then, weβve seen that we can apply both the quotient and chain rule to a problem involving the derivatives of trigonometric functions.